Physics, 13.12.2019 23:31 emurray

A2.0-kg silverware drawer does not slide readily. the owner gradually pulls with more and more force, and when the applied force reaches 9.0 n, the drawer suddenly opens, throwing all the utensils to the floor. what is the coefficient of static friction between the drawer and the cabinet?

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A 2.0-kg silverware drawer opens suddenly when the applied force reaches 9.0 N. The coefficient of static friction between the drawer and the cabinet is 0.46.

A drawer starts moving when the force applied is 9.0 N. At that moment, the force done by the person and the force of static friction are balanced. Thus the force of static friction (F) is 9.0 N.

We can calculate the coefficient of static friction (k) using the following expression.

$F = k \times N = k \times m \times g$

where,

N: normal forcem: mass of the drawerg: gravity (9.8 m/s²)

The coefficient of static friction is:

$k = \frac{F}{m \times g } = \frac{9.0N}{2.0kg \times (9.8m/s^{2} ) } = 0.46$

A 2.0-kg silverware drawer opens suddenly when the applied force reaches 9.0 N. The coefficient of static friction between the drawer and the cabinet is 0.46.

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Answer from: episodegirl903

0.45

Explanation:

First note that the body that causes the body to move is its moving force (Fm) which is 9.0N

Since the mass of the body is 2.0kg, the weight will be;

W= mg = 2×10

W= 20N

For static body, the frictional force (Ff) acting on the body is equal to the moving force (Fm) since both forces acts along the horizontal on the body.

Ff = Fm = 9.0N

The normal reaction (R) on the body will also be equal to its weight(W) since weight acts downwards and the reaction acts in the opposite direction (upwards).

R = W = 20N

Ff = nR taking 'n' as coefficient of static friction between the drawer and the cabinet.

9.0 = 20n

n = 9/20

n = 0.45

Answer from: Quest

Ithink it is letter b, i may be wrong but i think that is the answer

Answer from: Quest

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

$a = 32 ft/s^2$

so as we know that

$a = \frac{v^2}{r}$

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

$a_c = 32 ft/s^2$

$a_t = 0$

now radius of curvature is given as

$r = \frac{v^2}{a_c}$

$r = \frac{150^2}{32}$

$r = 703.125ft$

so the radius of curvature just after it is dropped is 703.125 ft

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